...函数,其中f,g均为可微函数。证明du/dy=g(z)du/dx.
发布网友
发布时间：2024-10-24 10:48
共1个回答
热心网友
时间：2024-10-27 14:25
=>dz/dx=1/(1-yg'(z))
dz/dy=g(z)+yg'(z)dz/dy
=>dz/dy=g(z)/(1-yg'(z))
du/dy=df/dy=(df/dz)·(dz/dy)
=g(z)(df/dz)/(1-yg'(z))
du/dx=df/dx=(df/dz)·(dz/dx)
=(df/dz)/(1-yg'(z))
∴du/dy=g(z)du/dx
